Question

.researcher suspected that the number of between meal snacks eaten by students in a day during final examinations
might depend on the number of tests a student had to take on that day. The accompanying table shows joint probabilities,
estimated from a survey.
Number of
No of snacks Y No of test
0 1 2 3
0
1 0.07 0.09 0.06 0.01
2   0.07 0.06 0.07 0.01
3   0.06 0.07 0.14 0.03
4   0.02 0.04 0.16 0.04
.
I. . Find probability of mean and variance of 3X-2Y
II. . Find the covariance between X and Y.
B. Find mean and variance of distribution by using M.G.F technique.(1− ?)^×(p)

Answer 1

i.

Joint distribution of (X,Y):

P(X=0,Y=0)=0.07, P(X=0,Y=1)=0.09, P(X=0,Y=2)=0.06, P(X=0,Y=3)=0.01,

P(X=1,Y=0)=0.07, P(X=1,Y=1)=0.06, P(X=1,Y=2)=0.07, P(X=1,Y=3)=0.01,

P(X=2,Y=0)=0.06, P(X=2,Y=1)=0.07, P(X=2,Y=2)=0.14, P(X=2,Y=3)=0.03,

P(X=3,Y=0)=0.02, P(X=3,Y=1)=0.04, P(X=3,Y=2)=0.16, P(X=3,Y=3)=0.04.

Let Z=3X-2Y, the pmf of Z is

P(Z=-6)=P(X=0,Y=3)=0.01;

P(Z=-4)=P(X=0,Y=2)=0.06;

P(Z=-3)=P(X=1,Y=3)=0.01;

P(Z=-2)=P(X=0,Y=1)=0.09;

P(Z=-1)=P(X=1,Y=2)=0.07;

P(Z=0)=P(X=2,Y=3)+P(X=0,Y=0)=0.03+0.07=0.10

P(Z=1)=P(X=1,Y=1)=0.06;

P(Z=2)=P(X=2,Y=2)=0.14;

P(Z=3)=P(X=3,Y=3)+P(X=1,Y=0)=0.04+0.07=0.11;

P(Z=4)=P(X=2,Y=1)=0.07;

P(Z=5)=P(X=3,Y=2)=0.16;

P(Z=6)=P(X=2,Y=0)=0.06;

P(Z=7)=P(X=3,Y=1)=0.04;

P(Z=9)=P(X=3,Y=0)=0.02.

E(Z)=-6*0.01+(-4)*0.06+(-3)*0.01+(-2)*0.09+(-1)*0.07+0*0.10+1*0.06+2*0.14+3*0.11+4*0.07+5*0.16+6*0.06+7*0.04+9*0.02=2.11.

ii.

E(X)=0*(0.07+0.09+0.06+0.01)+1*(0.07+0.06+0.07+0.01)+2*(0.06+0.07+0.14+0.03)+3*(0.02+0.04+0.16+0.04)

=1.59

E(Y)=0*(0.07+0.07+0.06+0.02)+1*(0.09+0.06+0.07+0.04)+2*(0.06+0.07+0.14+0.16)+3*(0.01+0.01+0.03+0.04)

=1.39

E(XY)=0*0.07+0*0.09+0*0.06+0*0.01+0*0.07+1*1*0.06+1*2*0.07+1*3*0.01+2*0*0.06+2*1*0.07+2*2*0.14+2*3*0.03+3*0*0.02+3*1*0.04+3*2*0.16+3*3*0.04=2.55

Cov(X,Y)=E(XY)-E(X)E(Y)=2.55-1.59*1.39=0.3399.